Higher Prelim Topic Revision

13 The Circle

Points of intersection can be found by solving simultaneous equations. For a line intersecting a circle, first get the line's equation in a suitable form then solve through substitution.

\((1,2)\) and \((8,-5)\)

For more practice, try: 1, 3 and 5 from these past exam questions.

Finding points of contact for a line tangent to a circle, or showing that a line is a tangent to a circle, can be approached just as a special case of finding points of intersection. Follow the approach of the previous question, and look at how the resulting quadratic factorises to show tangency.

\((2,-1)\) and repeated factors therefore the line is a tangent to the circle.

For more practice, try: 2, 4 and 6 from these past exam questions.

Unlike finding a tangent to other curves earlier in the course, finding the gradient of a tangent to a circle will not involve differentiation. Instead, start by finding the gradient of the radius which meets the tangent at the point of contact, and consider its gradient might lead to the gradient of the tangent.

\(y=\frac{3}{2}x-2\) or \(2y=3x-4\)

For more practice, try: these past exam questions.

For (a) remember that the formula sheet states the connection between the equation of a circle and its radius and centre.

For (b) it would help to know the distance the point is from the centre of the circle.

Centre \((5,-3)\) and radius \(\sqrt{17}\).
\(\sqrt{41}>\sqrt{17}\) so the point lies outside the circle (or other valid approach).

For more practice, try: Zeta Higher Tetbook, Page 50, Exercise 3.4, Q1 parts (a), (b) and (c).

Using the distance formula to find the answer to part (a) will be needed for part (b). Consider the radii of each circle. What is their sum?

For part (c), the radii of the circle have already been found. The ratio of the radii is the ratio in which the point P divides the line between the centres. This skill was previously introduced in the Vectors topic.